Question

Find \(\vec{N}(t)\). $$ \vec{r}(t)=\langle a \cos t, a \sin t, b t\rangle ; \quad a>0 $$

Short answer

\(\vec{N}(t) = \langle -\cos t, -\sin t, 0 \rangle\).

Step-by-step solution

Find the velocity vector \(\vec{v}(t)\)

We find the velocity vector \(\vec{v}(t)\) by differentiating the position vector \(\vec{r}(t)\) with respect to time \(t\). Thus: \( \vec{v}(t) = \frac{d}{dt}\vec{r}(t) = \left\langle -a \sin t, a \cos t, b \right\rangle \).

Find the acceleration vector \(\vec{a}(t)\)

Differentiating the velocity vector \(\vec{v}(t)\) with respect to \(t\), we get the acceleration vector: \( \vec{a}(t) = \frac{d}{dt}\vec{v}(t) = \left\langle -a \cos t, -a \sin t, 0 \right\rangle \).

Compute the unit tangent vector \(\vec{T}(t)\)

The unit tangent vector \(\vec{T}(t)\) is found by normalizing the velocity vector \(\vec{v}(t)\): \( \vec{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|} = \frac{\left\langle -a \sin t, a \cos t, b \right\rangle}{\sqrt{a^2 + b^2}} \), where \(\|\vec{v}(t)\| = \sqrt{a^2 + b^2}\).

Find the derivative of the unit tangent vector \(\frac{d\vec{T}(t)}{dt}\)

Differentiate \(\vec{T}(t)\): \( \frac{d\vec{T}(t)}{dt} = \frac{d}{dt}\left(\frac{\left\langle -a \sin t, a \cos t, b \right\rangle}{\sqrt{a^2 + b^2}}\right) = \left\langle -\frac{a \cos t}{\sqrt{a^2 + b^2}}, -\frac{a \sin t}{\sqrt{a^2 + b^2}}, 0 \right\rangle \).

Calculate the magnitude of \(\frac{d\vec{T}(t)}{dt}\)

Calculate the magnitude of \(\frac{d\vec{T}(t)}{dt}\): \( \left\| \frac{d\vec{T}(t)}{dt} \right\| = \sqrt{\left(-\frac{a \cos t}{\sqrt{a^2 + b^2}}\right)^2 + \left(-\frac{a \sin t}{\sqrt{a^2 + b^2}}\right)^2} = \frac{a}{\sqrt{a^2 + b^2}} \).

Determine the unit normal vector \(\vec{N}(t)\)

The unit normal vector \(\vec{N}(t)\) can be determined by normalizing \(\frac{d\vec{T}(t)}{dt}\): \( \vec{N}(t) = \frac{\frac{d\vec{T}(t)}{dt}}{\left\| \frac{d\vec{T}(t)}{dt} \right\|} = \frac{1}{a} \cdot \left\langle -a \cos t, -a \sin t, 0 \right\rangle = \left\langle -\cos t, -\sin t, 0 \right\rangle \).

Key concepts

Velocity Vector

The velocity vector, often denoted as \( \vec{v}(t) \), is crucial in understanding the trajectory of a moving object. It represents the rate of change of the position vector with respect to time. To find the velocity vector, we differentiate the given position vector \( \vec{r}(t) \) with respect to time \( t \). In this scenario, the position vector \( \vec{r}(t) = \langle a \cos t, a \sin t, b t \rangle \) becomes the velocity vector \( \vec{v}(t) = \langle -a \sin t, a \cos t, b \rangle \) after differentiation. This computation highlights how the velocity vector is essentially the derivative of the position vector, providing insight into both the direction and speed of an object moving along a path.

Key points to remember:

• Differentiation of position vector yields the velocity vector.
• The velocity vector indicates the direction in which the object is moving at any point, along with the instantaneous speed.

Acceleration Vector

The acceleration vector \( \vec{a}(t) \) is derived from the velocity vector by taking a further derivative with respect to time. It provides the rate at which the velocity is changing. In the given problem, we differentiate the previously found velocity vector \( \vec{v}(t) = \langle -a \sin t, a \cos t, b \rangle \) to obtain the acceleration vector \( \vec{a}(t) = \langle -a \cos t, -a \sin t, 0 \rangle \).

This vector shows how the speed and direction are changing for an object in motion. The components of the acceleration vector can demonstrate circular motion characteristics or a consistent linear acceleration.

Key insights:

• Further differentiation of the velocity vector gives us the acceleration vector.
• Acceleration vector components convey changes in motion, both directionally and in magnitude.

Unit Tangent Vector

The unit tangent vector, denoted as \( \vec{T}(t) \), normalizes the velocity vector to a unit vector (a vector with a magnitude of 1). This vector points in the direction of the curve that the object is following at any given instant. To compute it, we take the velocity vector \( \vec{v}(t) = \langle -a \sin t, a \cos t, b \rangle \) and divide it by its magnitude \( \|\vec{v}(t)\| = \sqrt{a^2 + b^2} \). Hence, the unit tangent vector is \( \vec{T}(t) = \frac{\langle -a \sin t, a \cos t, b \rangle}{\sqrt{a^2 + b^2}} \).

This operation is essential for simplifying the analysis of the path's direction independent of its speed, leading to a clear understanding of the path's geometry.

• The unit tangent vector reveals the direction along the path.
• Normalizing helps separate directional analysis from speed analysis.

Derivative of Vectors

The process of finding the derivative of a vector, such as the unit tangent vector, plays a critical role in further understanding motion. By differentiating the unit tangent vector \( \vec{T}(t) \), we uncover information about how the path's direction changes over time. In our example, the derivative \( \frac{d\vec{T}(t)}{dt} \) is computed as \( \left\langle -\frac{a \cos t}{\sqrt{a^2 + b^2}}, -\frac{a \sin t}{\sqrt{a^2 + b^2}}, 0 \right\rangle \).
This derivative helps in understanding the curvature of the trajectory. Moreover, when we normalize this derivative, we find the unit normal vector \( \vec{N}(t) \), which is crucial for understanding how sharply a path curves at any point.

Essential takeaways:

• Vector derivatives help analyze directional changes in a path.
• The derivative of the unit tangent vector is pivotal for understanding the path's curvature.
• Normalization following differentiation reveals the unit normal vector for curve analysis.